SOLUTION: Given the digits 3,9,5,6& 2, howmany different 5 digit numbers reater than 41,000(using each digit once) can form?

Algebra ->  Permutations -> SOLUTION: Given the digits 3,9,5,6& 2, howmany different 5 digit numbers reater than 41,000(using each digit once) can form?      Log On


   

Question 69055: Given the digits 3,9,5,6& 2, howmany different 5 digit numbers reater than 41,000(using each digit once) can form?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The 1st digit can't be 2 or 3
So, it has to be 5,6,or 9, or 3 choices
One digit is used up and there are 4 choices for the next digit
There are 3 choices for the next
There are 2 choices for the next
And 1 choice for the last
multiply all the choices together to get all the possible combination
3%2A4%2A3%2A2%2A1+=+72 possible numbers greater than 41,000 using
each digit only once